# Solving matrix equation Ax=B

1. Dec 16, 2013

### T Niemistoinen

1. The problem statement, all variables and given/known data

You are given matrices A and B:

1 -2 1
A = 0 0 0
0 1 1

1
B = 0
2

Solve the equation Ax=B

2. Relevant equations

x=A^(-1)*B

3. The attempt at a solution

I carried out exactly the steps required for inverting a matrix. When got to th part where the determinant is calculated, I obtained a zero determinant. I thought this meant that the inverse matrix could not be calculated. This is also the result that an online matrix calculator gave for A. Hence I anwered "unsolvable" in the exam. Yet I was only given 1 out of 3 points.

What should I have done differently?

2. Dec 16, 2013

### Office_Shredder

Staff Emeritus
The fact that A has no inverse means that there exist some vectors B such that
Ax = B
is not solvable. However as long as B is in the image of A it is solvable. In particular for this problem notice that A always forces the second coordinate of Ax to be equal to zero - so if B had a nonzero second coordinate the problem would be unsolvable. Luckily B has zero for its second coordinate, so it will turn out that Ax=B is solvable in this case.

As far as figuring out what x can be equal to, you can write down a system of equations from Ax=B which you should be able to solve. Try that and show us what you get!

3. Dec 16, 2013

### Ray Vickson

For any nxn linear system, exactly one of the following statements is true: Ax = b has (i) exactly one solution; (ii) no solutions; or (iii) infinitely many solutions. You chose (ii), but without checking the details. The easiest way is to just write out the equations:
$$1x -2y + 1x = 1\\ 0x+0y+0z=0\\ 0x+1y+1z = 2$$
The second equation is redundant, so we just have two equations in three unknowns:
$$\begin{array}{rc} x - 2y + z & = 1\\ y + z &= 2 \end{array}$$

4. Dec 17, 2013

### T Niemistoinen

Thanks alot guys! That's actually all familiar stuff to me and it all makes sense now, but for some reason I just can't think straight when it comes to matrices. Thanks again!